\documentclass{article} \usepackage[nosection]{incertia} \pagestyle{fancy} \lhead{Will Song} \chead{Dirichlet's Theorem} \rhead{\today} \begin{document} \begin{df} The $n$-th \textbf{cyclotomic polynomial} $\Phi_n$ is defined to be $\Phi_n(X) = \prod_{\mathclap{\substack{\zeta^n = 1 \\ \zeta^j \neq 1 \\ 0 < j < n}}} (X - \zeta),$ where $\zeta$ runs over all the primitive $n$-th roots of unity. \end{df} \begin{prop} $\deg \Phi_n = \varphi(n)$. \end{prop} \begin{proof} Trivial. \end{proof} \begin{prop} $X^n - 1 = \prod_{d \mid n} \Phi_d(X)$. \end{prop} \begin{proof} This is a simple exercise in root counting. Recall that $X^n - 1 = \prod_{\zeta^n = 1} (X - \zeta)$. For each $\zeta$ such that $\zeta^n = 1$, we can define $\ord(\zeta)$ as the smallest $k$ such that $\zeta^k = 1$. Notice that $\ord(\zeta) \mid n$. The factor $X - \zeta$ occurs in precisely $\Phi_{\ord(\zeta)}(X)$ and nothing else. \end{proof} \begin{cor} $n = \sum_{d \mid n} \varphi(d)$. \end{cor} \begin{proof} Trivial. Just add the degrees. \end{proof} \begin{prop} $\Phi_n(X) \in \ZZ[X]$. \end{prop} \begin{proof} We proceed with induction. The base case $\Phi_1(X) = X - 1$ is clearly true. Now suppose for all $k < n$ we have that $\Phi_k(X) \in \ZZ[X]$. Define $P_n(X) = \prod_{\mathclap{\substack{d \mid n \\ d \neq n}}} \Phi_d(X).$ Clearly $P_n$ is monic. We can then use the division algorithm to find integer polynomials $Q(X), R(X)$ such that $X^n - 1 = P_n(X) Q(X) + R(X)$ and $\deg R < \deg P_n = n - \varphi(n)$ or $R(X) = 0$. By plugging in the roots of $P_n$, it is easy to see that $R(X)$ must have at least $n - \varphi(n)$ roots, so if $R$ is non-zero it must have degree at least $n - \varphi(n)$, which cannot happen. Thus $R(X) = 0$ and $X^n - 1 = P_n(X) Q(X)$, where $Q(X) \in \ZZ[X]$, but $X^n - 1 = P_n(X) \Phi_n(X)$, which gives us $Q(X) = \Phi_n(X)$ and $\Phi_n(X) \in \ZZ[X]$. \end{proof} \begin{prop} $\Phi_n(X)$ is irreducible. \end{prop} \begin{proof} We actually don't have enough tools to show this for the general case. It is easy to show that $\Phi_p(X + 1)$ is irreducible and thus $\Phi_p(X)$ is also irreducible when $p$ is prime though. Just compute $\Phi_p(X) = 1 + X + X^2 + \cdots + X^{p - 1}$ and apply Eisenstein at $X + 1$. \end{proof} We now take some inspiration from standard calculus, and state some properties that we will not prove. Note that these hold in $\RR[X]$ and $\ZZ[X]$, but more importantly, they are also true in $\FF_p[X]$. \begin{prop} $(f(x) + g(x))' = f'(x) + g'(x)$. \end{prop} \begin{prop} $(f(x) g(x))' = f'(x) g(x) + f(x) g'(x)$. \end{prop} We can now prove the following statement. \begin{prop} Let $P$ be a polynomial over either $\RR[X]$, $\QQ[X]$, $\ZZ[X]$, or $\FF_p[X]$. Then there exists a non-constant polynomial $m(X)$ such that $m(X)^2 \mid P(X)$ if and only if $\gcd(P(X), P'(X)) \neq 1$. \end{prop} \begin{proof} Suppose there exists such a $m(X)$ such that $m(X)^2 \mid P(X)$. Then $P(X) = m(X)^2 Q(X)$ and $P'(X) = 2 m(X) m'(X) Q(X) + m(X)^2 Q'(X)$, so $m(X) \mid \gcd(P(X), P'(X))$. For the other direction, $\gcd(P(X), P'(X)) \neq 1$ implies that they share a root $X_0$. We claim that $m(X) = X - X_0$ satisfies the double root property for $P(X)$. Suppose not. Then $P(X) = m(X) Q(X)$ and $Q(X_0) \neq 0$. Compute $P'(X) = m'(X) Q(X) + m(X) Q'(X)$. Note that $P'(X_0) = 0$ because $X_0$ is a root of $P'(X)$, but $P'(X_0) = m'(X_0) Q(X_0) + m(X_0) Q'(X_0) = 1$, which is a contradiction. \end{proof} \begin{cor} $X^n - 1$ only has roots of multiplicity $1$ in any of the above rings. \end{cor} \begin{prop} Let $m, n$ be two positive integers and $p$ a prime such that $p \nmid mn$. Then $\gcd(\Phi_n(X), \Phi_m(X)) = 1$ over $\FF_p[X]$. \end{prop} \begin{proof} By the previous proposition $X^{mn} - 1$ taken in $\FF_p[X]$ has no repeated factors. Suppose for the sake of contradiction that $\gcd(\Phi_n(X), \Phi_m(X)) = g(X) \neq 1$. Then $g(X)^2 \mid X^{mn} - 1$, which is not true. This is because $X^{mn} - 1 = \prod_{d \mid mn} \Phi_d(X)$ and $\Phi_n, \Phi_m$ both show up in this product. \end{proof} \begin{cor} Let $m, n$ be two positive integers and $p$ prime such that $p \nmid mn$. Then $\Phi_n(x), \Phi_m(x)$ cannot be both divisible by $p$ for the same value of $x$. \end{cor} \begin{proof} Suppose $\Phi_n(x) \equiv 0 \pmod{p}$ and $\Phi_m(x) \equiv 0 \pmod{p}$. Then $X - x$ is a factor of $\Phi_n(X)$ and $\Phi_m(X)$ in $\FF_p[X]$, which cannot happen. \end{proof} For a given prime $p$, we define the standard order $\ord_p(a)$ to be the least positive integer $k$ such that $a^k \equiv 1 \pmod{p}$. Below we give one of the most important theorems in the theory of cyclotomic polynomials. \begin{thm} Let $p$ be a prime. Then for all positive integers $n$ and integers $a$ such that $\gcd(n, p) = 1$, we have $p \mid \Phi_n(a) \iff \ord_p(a) = n$. \end{thm} \begin{proof} We proceed via induction on $n$. The base case $n = 1$ is trivial since $\Phi_1(x) = x - 1$ has a root at $x \equiv 1 \pmod{p}$. Now suppose that the hypothesis is true for all $k < n$. Suppose $a$ satisfies $\Phi_n(a) \equiv 0 \pmod{p}$. Recall that $a^n - 1 = \prod_{d \mid n} \Phi_d(a) \equiv 0 \pmod{p}$, so $a^n \equiv 1 \pmod{p}$. Now suppose that $\ord_p(a) = k \neq n$. By the induction hypothesis, we know that $\Phi_k(a) = 0$. We also have $k \mid n$, so $\gcd(k, p) = 1$ implies $p \nmid kn$, which contradicts proposition 12. Thus $\ord_p(a) = n$. Now suppose $\ord_p(a) = n$. Then $\prod_{d \mid n} \Phi_d(a) = a^n - 1 \equiv 0 \pmod{p}$. If for some $d < n$ we have $\Phi_d(a) \equiv 0 \pmod{p}$, we have $\ord_p(a) = d$ by the induction hypothesis so $\prod_{\mathclap{\substack{d \mid n \\ d \neq n}}} \Phi_d(a) \not \equiv 0 \pmod{p}$. Thus we necessarily have $\Phi_n(a) \equiv 0 \pmod{p}$. \end{proof} We can now finally prove the main theorem. \begin{thm}[Dirichlet] Let $a, n$ be two relatively prime positive integers. Then there exist infinitely many primes $p$ of the form $p \equiv a \pmod{n}$. \end{thm} Okay I lied. We actually can't prove that. We can only prove the special case of $a = 1$. \begin{cor}[Dirichlet] For any integer $n$ there exist infinitely many primes of the form $p \equiv 1 \pmod{n}$. \end{cor} We first prove the following lemmas. \begin{lem} Let $p$ be a prime not dividing $n$. If there exists an integer $a$ such that $p \mid \Phi_n(a)$, then $p \equiv 1 \pmod{n}$. \end{lem} \begin{proof} By theorem 13, we immediately know that $\ord_p(a) = n$, but this implies $n \mid p - 1$, or $p - 1 \equiv 0 \pmod{n}$, giving the result. \end{proof} \begin{lem}[Schur] Given an integer polynomial $f$, there exist infinitely many primes $p$ such that there exists an integer $a_p$ such that $p \mid f(a_p)$. \end{lem} \begin{proof} We can write $f(n) = n g(n) + f(0)$. If $f(0) = 0$, then the result is obvious. Now suppose $f(0) \neq 0$. Suppose without loss of generality that $\lim_{n \to \infty} f(n) = +\infty$. Let $f(0) = c$. That means there exists a sufficiently large $N$ such that $k > N \implies f(k) > |c|$. Pick any of these $k$. Compute $f(c^2 k!) = c^2 k! g(c^2 k!) + c = c (c k! g(c^2 k!) + 1).$ Then $c k! g(c^2 k!) + 1$ must be divisible by some prime. It is also clearly not divisible by any prime $p \leq k$ because $k! \equiv 0 \pmod{p}$. Thus this is divisible by some prime larger than $k$ so we can just take a sequence of sufficiently large $k$ to get our infinite primes. \end{proof} Now we can prove the corollary. \begin{proof} We can use lemma 17 to get infinitely many pairs $p, a_p$ such that $p \mid \Phi_n(a_p)$, and we immediately get $p \equiv 1 \pmod{n}$ by lemma 16. \end{proof} The above method can be strengthened into the following proposition, but we do not have the tools necessary for proving this, so we will just leave it at that. \begin{prop} For all integers $n$ and integers $a$ such that $a^2 \equiv 1 \pmod{n}$, there are infinitely many primes $p$ of the form $p \equiv a \pmod{n}$. \end{prop} \end{document}